A \"seconds\" pendulum is one that goes through its equilibriumposition once eac

Question

A "seconds" pendulum is one that goes through its equilibriumposition once each second. (The period of the pendulum is 2.000 s.)The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 mat Cambridge, England. What is the ratio of the free-fallaccelerations at these two locations? (Cambridge / Tokyo) = 1 + A "seconds" pendulum is one that goes through its equilibriumposition once each second. (The period of the pendulum is 2.000 s.)The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 mat Cambridge, England. What is the ratio of the free-fallaccelerations at these two locations? (Cambridge / Tokyo) = 1 +

Explanation / Answer

Modified: ---------- we know T = 2 [ L / g ] for seconds pendulum time period T = 2 s i.e., period T = constant L g ( gc / g t ) =[ L c / L t ]                 = [ 0.9942 m / 0.9927 ]                 = 1.001511                  =1 + 0.001511

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